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Yes because 21812/2 = 10906

Q: Is 2 divisible by 21 812?

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If its divisible by 5 AND 2 it must be divisible by 10 So you just have to pick the only number between 21 and 39 that's divisible by 10

21 is not divisible by 2 nor 5, but it is divisible by 3. To be divisible by 2 the last digit must be even (one of {0, 2, 4, 6, 8}; the last digit 1 is not one of these, thus 21 is not divisible by 2 To be divisible by 3 sum the digits and if this total is divisible by 3, then so is the original number; the test can be repeated on the sum, so by repeatedly summing the digits of the totals until a single digit remains only if that single digit is 3, 6 or 9 is the original number divisible by 3. 21 → 2 + 1 = 3 which is divisible by 3 so 21 is divisible by 3. To be divisible by 5, the last digit must be 0 or 5. The last digit of 21 is 1 which is not 0 nor 5, thus 21 is not divisible by 5.

812/2 = 406 406/2 = 203 207/7 = 29 2*2*7*29 = 28*29 = 812

42 can be divided by 2 once. The result, 21, is not divisible by 2.

The divisibility rule for 3 is as follows: If the sum of the digits of a number is divisible by 3 then the number is divisible by 3. For example, the number 21 is divisible by 3 since 2+1=3, and 3 is divisible by 3. The number 49 is not divisible by 3 since 4+9=13, and 13 is not divisible by 3.

Related questions

2 and 4

812 is not divisible by 9. The divisors of 812 are 1, 2, 4, 7, 14, 28, 29, 58, 116, 203, 406, and 812.

No. No number can be divisible by a number greater than it. 812 is greater than 6 so 6 cannot be divisible by 812.

No. 812 is not evenly divisible by 63.

45 yes, 812 no.

812 is not divisible by 9. The divisors of 812 are 1, 2, 4, 7, 14, 28, 29, 58, 116, 203, 406, and 812.

2 , 4 and 7 if you are listing through ten

103

No, because 21 is not divisible by 2.

The LCM of 2 and 812 is 812. The LCM of 2, 8 and 12 is 24.

Yes, but only as a decimal. It does not divide evenly.

yes 9 x 812 = 7,308